3.98 \(\int \frac{a+b \sinh ^{-1}(c x)}{x^3 (\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=162 \[ \frac{3 b c^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{3 b c^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi c^2 x^2+\pi }}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{3 c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2}}+\frac{b c^2 \tan ^{-1}(c x)}{\pi ^{3/2}}-\frac{b c}{2 \pi ^{3/2} x} \]

[Out]

-(b*c)/(2*Pi^(3/2)*x) - (3*c^2*(a + b*ArcSinh[c*x]))/(2*Pi*Sqrt[Pi + c^2*Pi*x^2]) - (a + b*ArcSinh[c*x])/(2*Pi
*x^2*Sqrt[Pi + c^2*Pi*x^2]) + (b*c^2*ArcTan[c*x])/Pi^(3/2) + (3*c^2*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x
]])/Pi^(3/2) + (3*b*c^2*PolyLog[2, -E^ArcSinh[c*x]])/(2*Pi^(3/2)) - (3*b*c^2*PolyLog[2, E^ArcSinh[c*x]])/(2*Pi
^(3/2))

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Rubi [A]  time = 0.350337, antiderivative size = 212, normalized size of antiderivative = 1.31, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5747, 5755, 5760, 4182, 2279, 2391, 203, 325} \[ \frac{3 b c^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{3 b c^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi c^2 x^2+\pi }}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{3 c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2}}-\frac{b c \sqrt{c^2 x^2+1}}{2 \pi x \sqrt{\pi c^2 x^2+\pi }}+\frac{b c^2 \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{\pi \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x^3*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(2*Pi*x*Sqrt[Pi + c^2*Pi*x^2]) - (3*c^2*(a + b*ArcSinh[c*x]))/(2*Pi*Sqrt[Pi + c^2*Pi*
x^2]) - (a + b*ArcSinh[c*x])/(2*Pi*x^2*Sqrt[Pi + c^2*Pi*x^2]) + (b*c^2*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(Pi*Sqrt
[Pi + c^2*Pi*x^2]) + (3*c^2*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Pi^(3/2) + (3*b*c^2*PolyLog[2, -E^Ar
cSinh[c*x]])/(2*Pi^(3/2)) - (3*b*c^2*PolyLog[2, E^ArcSinh[c*x]])/(2*Pi^(3/2))

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5755

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp
[((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[
c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}-\frac{1}{2} \left (3 c^2\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx+\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 \pi \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{2 \pi x \sqrt{\pi +c^2 \pi x^2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi +c^2 \pi x^2}}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}-\frac{\left (3 c^2\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{\pi +c^2 \pi x^2}} \, dx}{2 \pi }-\frac{\left (b c^3 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 \pi \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (3 b c^3 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 \pi \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{2 \pi x \sqrt{\pi +c^2 \pi x^2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi +c^2 \pi x^2}}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{b c^2 \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt{\pi +c^2 \pi x^2}}-\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{3/2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{2 \pi x \sqrt{\pi +c^2 \pi x^2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi +c^2 \pi x^2}}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{b c^2 \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt{\pi +c^2 \pi x^2}}+\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac{\left (3 b c^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{3/2}}-\frac{\left (3 b c^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \pi ^{3/2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{2 \pi x \sqrt{\pi +c^2 \pi x^2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi +c^2 \pi x^2}}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{b c^2 \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt{\pi +c^2 \pi x^2}}+\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac{\left (3 b c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{\left (3 b c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}\\ &=-\frac{b c \sqrt{1+c^2 x^2}}{2 \pi x \sqrt{\pi +c^2 \pi x^2}}-\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi \sqrt{\pi +c^2 \pi x^2}}-\frac{a+b \sinh ^{-1}(c x)}{2 \pi x^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{b c^2 \sqrt{1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt{\pi +c^2 \pi x^2}}+\frac{3 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac{3 b c^2 \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}-\frac{3 b c^2 \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \pi ^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.94736, size = 269, normalized size = 1.66 \[ \frac{-12 b c^2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+12 b c^2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-\frac{8 a c^2}{\sqrt{c^2 x^2+1}}-\frac{4 a \sqrt{c^2 x^2+1}}{x^2}+12 a c^2 \log \left (\pi \left (\sqrt{c^2 x^2+1}+1\right )\right )-12 a c^2 \log (x)-\frac{8 b c^2 \sinh ^{-1}(c x)}{\sqrt{c^2 x^2+1}}-12 b c^2 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+12 b c^2 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 b c^2 \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-2 b c^2 \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-b c^2 \sinh ^{-1}(c x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )-b c^2 \sinh ^{-1}(c x) \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(c x)\right )+16 b c^2 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )}{8 \pi ^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x^3*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

((-8*a*c^2)/Sqrt[1 + c^2*x^2] - (4*a*Sqrt[1 + c^2*x^2])/x^2 - (8*b*c^2*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] + 16*b*
c^2*ArcTan[Tanh[ArcSinh[c*x]/2]] - 2*b*c^2*Coth[ArcSinh[c*x]/2] - b*c^2*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 -
12*b*c^2*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 12*b*c^2*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] - 12*a*c^2
*Log[x] + 12*a*c^2*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] - 12*b*c^2*PolyLog[2, -E^(-ArcSinh[c*x])] + 12*b*c^2*PolyLo
g[2, E^(-ArcSinh[c*x])] - b*c^2*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*b*c^2*Tanh[ArcSinh[c*x]/2])/(8*Pi^(3/2
))

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Maple [A]  time = 0.177, size = 234, normalized size = 1.4 \begin{align*} -{\frac{a}{2\,\pi \,{x}^{2}}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{3\,a{c}^{2}}{2\,\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}+{\frac{3\,a{c}^{2}}{2\,{\pi }^{3/2}}{\it Artanh} \left ({\sqrt{\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}} \right ) }-{\frac{3\,b{\it Arcsinh} \left ( cx \right ){c}^{2}}{2\,{\pi }^{3/2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{bc}{2\,{\pi }^{3/2}x}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{2\,{\pi }^{3/2}{x}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+2\,{\frac{b{c}^{2}\arctan \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{{\pi }^{3/2}}}+{\frac{3\,b{c}^{2}}{2\,{\pi }^{3/2}}{\it dilog} \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{3\,b{c}^{2}}{2\,{\pi }^{3/2}}{\it dilog} \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{3\,b{\it Arcsinh} \left ( cx \right ){c}^{2}}{2\,{\pi }^{3/2}}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^3/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

-1/2*a/Pi/x^2/(Pi*c^2*x^2+Pi)^(1/2)-3/2*a*c^2/Pi/(Pi*c^2*x^2+Pi)^(1/2)+3/2*a*c^2/Pi^(3/2)*arctanh(Pi^(1/2)/(Pi
*c^2*x^2+Pi)^(1/2))-3/2*b/Pi^(3/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^2-1/2*b*c/Pi^(3/2)/x-1/2*b/Pi^(3/2)/(c^2*x
^2+1)^(1/2)/x^2*arcsinh(c*x)+2*b*c^2/Pi^(3/2)*arctan(c*x+(c^2*x^2+1)^(1/2))+3/2*b*c^2/Pi^(3/2)*dilog(c*x+(c^2*
x^2+1)^(1/2))+3/2*b*c^2/Pi^(3/2)*dilog(1+c*x+(c^2*x^2+1)^(1/2))+3/2*b*c^2/Pi^(3/2)*arcsinh(c*x)*ln(1+c*x+(c^2*
x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (\frac{3 \, c^{2} \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right )}{\pi ^{\frac{3}{2}}} - \frac{3 \, c^{2}}{\pi \sqrt{\pi + \pi c^{2} x^{2}}} - \frac{1}{\pi \sqrt{\pi + \pi c^{2} x^{2}} x^{2}}\right )} a + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^3/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxima")

[Out]

1/2*(3*c^2*arcsinh(1/(sqrt(c^2)*abs(x)))/pi^(3/2) - 3*c^2/(pi*sqrt(pi + pi*c^2*x^2)) - 1/(pi*sqrt(pi + pi*c^2*
x^2)*x^2))*a + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/((pi + pi*c^2*x^2)^(3/2)*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\pi + \pi c^{2} x^{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}}{\pi ^{2} c^{4} x^{7} + 2 \, \pi ^{2} c^{2} x^{5} + \pi ^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^3/(pi*c^2*x^2+pi)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^7 + 2*pi^2*c^2*x^5 + pi^2*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{2} x^{5} \sqrt{c^{2} x^{2} + 1} + x^{3} \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{5} \sqrt{c^{2} x^{2} + 1} + x^{3} \sqrt{c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**3/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a/(c**2*x**5*sqrt(c**2*x**2 + 1) + x**3*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(c**2*x**5*
sqrt(c**2*x**2 + 1) + x**3*sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^3/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((pi + pi*c^2*x^2)^(3/2)*x^3), x)